Integrals — Class 12 Maths Solution

: Let$I = \int {\cfrac{x}{{9 - 4{x^2}}}dx}$
Put $9 - 4{x^2} = t$ $\Rightarrow$ $- 8x\,dx = dt$

$\therefore$ $I = - \cfrac{1}{8}\int {\cfrac{{dt}}{t}} = - \cfrac{1}{8}\log \left| t \right| + C = \cfrac{1}{8}\log \cfrac{1}{{\left| t \right|}} + C$

$= \cfrac{1}{8}\log \cfrac{1}{{\left| {9 - 4{x^2}} \right|}} + C$